In Shreve II, exercise 1.8, he walks the reader through proving the derivative of a moment-generating function $\phi$ is equal to the expectation $\mathrm{E}[Xe^{tX}]$; i.e., $$ \phi^\prime(t) = \mathrm{E}[Xe^{tX}]. $$ He does this first assuming $X$ is nonnegative and asks the reader to use the dominated congergence theorem along with the mean value theorem, then asks the reader to prove it for an integrable $X$ by considering the positive and negative parts.

My question is, is all this machinery necessary for this exercise? Assuming $X$ has a density function $f$, I would prove this result by $$ \phi^\prime(t) = \frac{d}{dt}\int_{0}^\infty e^{tx}f(x) dx = \int_{0}^\infty \frac{d}{dt}e^{tx}f(x) dx = \int_{0}^\infty xe^{tx}f(x) dx = \mathrm{E}[Xe^{tX}], $$ where I guess I would have to justify moving the derivative inside the integral (is it enough that $e^{tx}$ is continuous in $t$?)

So, is this more rigorous proof in Shreve just to show it holds if $X$ doesn't have a density? Or perhaps just to practice using the dominated convergence theorem? I wish authors would state the spirit behind their exercises...

  • $\begingroup$ I have two questions: 1) Is X an absolutely continues random variable? 2) What conditions need to be $\frac{d}{dt}\int=\int\frac{d}{dt}$ $\endgroup$ – user16891 Jul 21 '15 at 17:48

As Sherev has said, first let $\varphi(t)=E\left[e^{tX}\right]$ then $$\varphi '(t)=\underset{s\to t}{\mathop{\lim }}\,\frac{\varphi (t)-\varphi (s)}{t-s}=\underset{s\to t}{\mathop{\lim }}\,\frac{E[{{e}^{tX}}]-E[{{e}^{sX}}]}{t-s}=\underset{s\to t}{\mathop{\lim }}\,E\left[ \frac{{{e}^{tX}}-{{e}^{sX}}}{t-s} \right]$$ Sherev continue ,we can choose a sequence of numbers $\{s_n\}_{n=1}^{\infty}$ that converges to $t$ and compute $$\underset{s_n\to t}{\mathop{\lim }}\,E\left[ \frac{{{e}^{tX}}-{{e}^{s_nX}}}{t-s_n} \right]$$ where now we are taking a limit of the expectations of the sequence of random variables. $$Y_n=\frac{{{e}^{tX}}-{{e}^{s_nX}}}{t-s_n}$$ he fixes $\omega\in \Omega$ and by application of Mean Value Theorem concludes $$e^{tX(\omega)}-e^{s_nX(\omega)}=(t-s_n)X(\omega)e^{\theta(\omega )X(\omega)}$$ where $\theta(\omega)$ is a number depending on $\omega$ such that $s_n\leq\theta(\omega)\leq t$.

$$|Y_n|=\left|\frac{e^{tX(\omega)}-e^{s_nX(\omega)}}{t-s_n}\right|\leq X(\omega) e^{\theta(\omega )X(\omega)}\leq X(\omega) e^{2t\,X(\omega)}$$ So by the Dominated Convergence Theorem we have $$\varphi'(t)=\underset{n\to \infty }{\mathop{\lim }}\,E[{{Y}_{n}}]=E[\underset{n\to \infty }{\mathop{\lim }}\,{{Y}_{n}}]=E[X{{e}^{tX}}]$$

  • $\begingroup$ Thansk for the solution, but I was just wondering why Shreve chose this general route instead of appealing to the much simpler case of continuous random variables. I can only guess that it was to show an application of the relevant theorems and to show the result holds for a general random variable (not necessarily a continuous one). $\endgroup$ – bcf Jul 21 '15 at 19:56
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    $\begingroup$ Actually, how did you obtain the last ineqaility $Xe^{\theta X} \leq Xe^{tX}$? It seems like restricting $s_n \leq t$ may not allow us to assume the equality $\lim_{n \to \infty} \mathrm{E}[Y_n] = \phi^\prime(t)$. OTOH we may still use Dom. Conv. since $|Y_n| \leq Xe^{2t X}$ for all $n$ large enough. $\endgroup$ – bcf Jul 21 '15 at 20:05
  • $\begingroup$ @user16891 How did you know that $\mathbb{E} Xe^{2tX}$ is bounded? $\endgroup$ – metricspace Oct 27 '17 at 4:22

The whole machinery is needed to interchange the integral sign and the derivative.


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